Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $r = \dfrac{-5q^2 - 25q - 20}{3q^2 + 33q + 84} \times \dfrac{q + 7}{q + 6} $
First factor out any common factors. $r = \dfrac{-5(q^2 + 5q + 4)}{3(q^2 + 11q + 28)} \times \dfrac{q + 7}{q + 6} $ Then factor the quadratic expressions. $r = \dfrac {-5(q + 4)(q + 1)} {3(q + 4)(q + 7)} \times \dfrac {q + 7} {q + 6} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac { -5(q + 4)(q + 1) \times (q + 7)} { 3(q + 4)(q + 7) \times (q + 6)} $ $r = \dfrac {-5(q + 4)(q + 1)(q + 7)} {3(q + 4)(q + 7)(q + 6)} $ Notice that $(q + 4)$ and $(q + 7)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac {-5\cancel{(q + 4)}(q + 1)(q + 7)} {3\cancel{(q + 4)}(q + 7)(q + 6)} $ We are dividing by $q + 4$ , so $q + 4 \neq 0$ Therefore, $q \neq -4$ $r = \dfrac {-5\cancel{(q + 4)}(q + 1)\cancel{(q + 7)}} {3\cancel{(q + 4)}\cancel{(q + 7)}(q + 6)} $ We are dividing by $q + 7$ , so $q + 7 \neq 0$ Therefore, $q \neq -7$ $r = \dfrac {-5(q + 1)} {3(q + 6)} $ $ r = \dfrac{-5(q + 1)}{3(q + 6)}; q \neq -4; q \neq -7 $